Number of points where the function f(x) = maximum ( 2x - x^2 ,2 … - Vidyadip

MCQ

Number of points where the function $f(x) = $ maximum $\left( {\sqrt {2x - {x^2}} ,2 - x} \right)$ is non differentiable is

A
$0$
B
$1$
✓
$2$
D
$3$

✓

Answer

Correct option: C.

$2$

c
at $x = 1 , 2$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

If $y = \sin (2{\sin ^{ - 1}}x),$ then ${{dy} \over {dx}} = $

Let $f(x) = \cos px + \sin x$ be periodic, then $p$ must be

Let $\vec \lambda = x\vec a + y\vec b + z\vec c$ and $\vec \lambda .(\vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a) = 2(x + y + z)$ (where $x + y + z \neq 0)$ then $\left[ {\vec a\,\,\vec b\,\,\vec c} \right]$ is

The value of $\mathop {\lim }\limits_{x \to 7} \frac{{2 - \sqrt {x - 3} }}{{{x^2} - 49}}$ is

If $\sin \left(\frac{y}{x}\right)=\log _0|x|+\frac{\alpha}{2}$ is the solution of the differential equation $x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x$ and $y(1)=\frac{\pi}{3}$, then $\alpha^2$ is equal to

$\lim \limits_{x \rightarrow 0} \frac{x\left(e^{\frac{\left(\sqrt{1+x^{2}+x^{4}}-1\right)}{x}}-1\right)}{\sqrt{1+x^{2}+x^{4}}-1}$ is equal to

$\int_{}^{} {{e^{{{\cos }^2}x}}\sin 2x\;dx = } $

Choose the correct number of ways in which $15$ different books can be divided into five heaps of equal number of books

If $x = a \sin \theta $ and $y = b$ $\cos \theta ,$ then ${{{d^2}y} \over {d{x^2}}}$ is

The equation $\arg \left(\frac{\mathrm{z}-1}{\mathrm{z}+1}\right)=\frac{\pi}{4}$ represents a circle with: